InterviewSolution
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InterviewSolution
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`8` mol of gas `AB_(3)` are introduced into a `1.0 dm^(3)` vessel. It dissociates as `2AB_(3)(g) hArr A_(2)(g)+3B_(2)(g)` At equilibrium, 2 mol of `A_(2)` is found to be present. The equilibrium constant for the reaction isA. `72 mol^(2)L^(-2)`B. `36 mol^(2)L^(-2)`C. `3 mol^(2)L^(-2)`D. `27 mol^(2)L^(-2)` |
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Answer» `{:(,2AB_(3(g)),hArr,A_(2(g)),+,3B_(2(g))),("At t=0",8,,0,,0),("At eqm.",8-a,,a//2,,3a//2):}` Thus `K_(c)=([A_(2)][B_(2)]^(3))/([AB_(3)]^(2))`, Also `(a)/(2)=2` `:. A=4` and `[AB_(3)]=(4)/(1)`, `[A_(2)]=(2)/(1)` and `[B_(2)]=(6)/(1)` `=(2xx6^(3))/(4^(2))=27 mol^(2)litre^(2-)` |
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